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State two causes of harmonic distortion in transistor amplifiers.(2 marks)craft1 electrical June/July 2020
Nonlinearity of the transistor: Transistors have a nonlinear current-voltage characteristic, which means that the current flowing through the transistor is not directly proportional to the voltage applied to it. This nonlinearity can cause harmonic distortion in the output signal. Saturation of theRead more
Define harmonic distortion with reference to audio amplifiers.(2 marks)craft1 electrical June/July 2020
Harmonic distortion is a type of distortion that occurs in audio amplifiers when the output signal contains additional frequency components that are not present in the input signal. These additional frequency components, called harmonics, are multiples of the fundamental frequency of the input signaRead more
Harmonic distortion is a type of distortion that occurs in audio amplifiers when the output signal contains additional frequency components that are not present in the input signal. These additional frequency components, called harmonics, are multiples of the fundamental frequency of the input signal and can cause the output signal to deviate from the original waveform. Harmonic distortion can be caused by a variety of factors, including the nonlinear behavior of the amplifier itself, the load being driven by the amplifier, and the presence of noise or other interference in the system. Harmonic distortion can cause the sound produced by the amplifier to be harsh or unpleasant, and it can also affect the accuracy and fidelity of the reproduced audio.
See lessList two: examples of secondary cells (2 marks)craft1 electrical June/July 2020
Lead-acid batteries: These are commonly used in vehicles and are made up of lead-based electrodes and a sulfuric acid electrolyte. Nickel-cadmium batteries: These are commonly used in portable electronic devices and are made up of nickel and cadmium electrodes and a potassium hydroxide electrolyte.Read more
List two:properties of primary cells: (2 marks)craft1 electrical June/July 2020
Primary cells are self-discharging, meaning that they cannot be recharged and must be replaced when the chemical reaction that generates the electrical current is complete. Primary cells have a finite lifespan, and their capacity (the amount of electrical energy they can produce) decreases as they aRead more
State two; advantages of parallel connection of electric loads.(2 marks)craft1 electrical June/July 2020
If one load in a parallel circuit fails or is disconnected, the other loads will continue to function because they are connected to the power source through separate paths. The voltage across each load in a parallel circuit is the same, so the loads can operate at their optimal voltage and functionRead more
State two; a (1) disadvantages of series connection of electric loads.(2 marks)craft1 electrical June/July 2020
If one load in the series circuit fails or is disconnected, the entire circuit will be interrupted and all of the loads will stop functioning. The voltage drop across each load in a series circuit is the same, so the loads may not operate at their optimal voltage and may not function properly. The tRead more
An electrical appliance is rated 2 kW and it is operated for 30 minutes. Determine the amount of heat produced. (3 marks)craft1 electrical June/July 2020
To find the amount of heat produced by the appliance, we can use the formula Q = P * t, where Q is the amount of heat produced, P is the power of the appliance, and t is the time it is operated. Plugging in the values given, we get Q = 2 kW * 30 minutes = 60 kJ. Note that the power is given in kilowRead more
To find the amount of heat produced by the appliance, we can use the formula Q = P * t, where Q is the amount of heat produced, P is the power of the appliance, and t is the time it is operated. Plugging in the values given, we get Q = 2 kW * 30 minutes = 60 kJ. Note that the power is given in kilowatts and the time is given in minutes, so we will need to convert the power to kilojoules per minute and the time to minutes to get the heat in kilojoules. Since 1 kW = 1000 J/s, 1 kW = 1000 J/s * 1 min/60 s = 16.67 J/s * 1 min = 16.67 J/min. Therefore, 2 kW = 2 * 16.67 J/min = 33.3 J/min. Thus, the amount of heat produced is 60 kJ.
See lessState two practical applications of heating effect of electric current.(2 marks)craft1 electrical June/July 2020
Electric ovens: Electric ovens use heating elements, which are metal wires that are heated by the flow of electric current through them. This heat is then used to cook or bake food. Soldering irons: Soldering irons are used to join two metal pieces together by melting a small amount of metal (solderRead more
An air cored toroidal coil has 3000 turns and carries current 0.1 A. the length of the magnetic circuit is 30 cm and the cross-sectional area of the coil is 4cm^2. Determine the: (i) magnetic field strength; (ii) magnetic flux density. (6 marks)craft1 electrical June/July 2020
i. To find the magnetic field strength, we can use the formula B = u0 * n * I / l, where B is the magnetic field strength, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, and l is the length of the magnetic circuit. Plugging in the values givRead more
i. To find the magnetic field strength, we can use the formula B = u0 * n * I / l, where B is the magnetic field strength, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / 30 cm = 4.210^-4 T.
ii. To find the magnetic flux density, we can use the formula B = u0 * n * I / (A * l), where B is the magnetic flux density, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, A is the cross-sectional area of the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / (4 cm^2 * 30 cm) = 3.110^-3 T.
See lessDefine the following terms as used in magnetism: (i) reluctance; (ii) hysteresis. (5 marks)craft1 electrical June/July 2020
(i) Reluctance is a measure of the resistance of a material to being magnetized. It is defined as the ratio of the magnetomotive force (MMF) required to produce a certain level of magnetization in the material to the resulting magnetization. The unit of reluctance is the ampere-turn per weber (At/WbRead more
(i) Reluctance is a measure of the resistance of a material to being magnetized. It is defined as the ratio of the magnetomotive force (MMF) required to produce a certain level of magnetization in the material to the resulting magnetization. The unit of reluctance is the ampere-turn per weber (At/Wb).
(ii) Hysteresis is the lag in the response of a material to changes in an applied magnetic field. It is typically observed as a loop on a graph of magnetization versus applied magnetic field, and it is a measure of the energy loss associated with reversing the magnetization of the material. In a material with hysteresis, it takes more energy to magnetize the material in one direction than it does to demagnetize it, and vice versa. This phenomenon is important in the design of devices such as transformers and motors, as it can result in losses due to the hysteresis loop.
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